Chapter One
Exercise
Write the electronic configuration of the following elements;
1S 2S
2P 3S 3P 4S
3D 4P 5S 4D 5P 6S
4F 5D 6P 7S 5F 6D 7P 8S
Ge atomic number 32
1S² 2S² 2P⁶ 3S² 3P⁶ 4S² 3D¹⁰ 4P²
Mn atomic number 25
1S² 2S² 2P⁶ 3S² 3P⁶ 4S² 3D⁵
Pb atomic number 82
1S² 2S² 2P⁶ 3S² 3P⁶ 4S² 3D¹⁰ 4P⁶ 5S² 4D¹⁰ 5P⁶ 6S² 4F¹⁴ 5D¹⁰ 6P⁶ 7S² 5F⁴
Mg atomic number 12
1S² 2S² 2P⁶ 3S²
P atomic number 15
1S² 2S² 2P⁶ 3S² 3P³
Ni atomic number 28
1S² 2S² 2P⁶ 3S² 3P⁶ 4S² 3D⁸
Cr atomic number 24
1S² 2S² 2P⁶ 3S² 3P⁶ 4S² 3D⁴
Cu atomic number 29
1S² 2S² 2P⁶ 3S² 3P⁶ 4S² 3D⁹
Al atomic number 13
1S² 2S² 2P⁶ 3S² 3P¹
Exercise
Question one
State two factors which determine the electrical conductivity of a material and hence discuss how
these factors are influenced by temperature to affect the electrical conductivity behavior of;
Answer; The number of free charge carriers and mobility of free charge carriers.
(i). super ionic materials [2 marks] In these materials, the electrical conductivity is primarily
determined by the concentration of mobile ions and their mobility. As the temperature increases,
the mobility of the ions increases, leading to an increase in the electrical conductivity.
(ii). Metals [2 marks] Metals are materials that have a high density of free electrons, which are
responsible for their high electrical conductivity. As the temperature increases, the average
thermal energy of the electrons increases, leading to an increase in their mobility.
(iii). Insulators [2 marks] Insulators have very few free charge carriers and, therefore, low
electrical conductivity. As the temperature increases, more electrons may be thermally excited
from the valence band to the conduction band, creating additional free charge carriers and
increasing the electrical conductivity.
(iv). Semiconductors [2 marks]: Semiconductors have a moderate number of free charge carriers
and intermediate electrical conductivity. The electrical conductivity of semiconductors is
determined by their bandgap energy, which defines the energy required to excite an electron
from the valence band to the conduction band. As the temperature increases, more electrons may
be thermally excited across the bandgap, leading to an increase in electrical conductivity.
Question two
(i). Define clearly from first principles what is electrical conductivity; give its formula and units
[3 marks]
Electrical conductivity is a measure of a material's ability to conduct an electrical current.
𝐼
(𝜎 = 𝐴×𝐿); units (siemens per meter (S/m) or ohm per meter (𝛺/m).
(ii). Explain why heat and current conduction electrons are only available in the conduction band
whereas bonding electrons reside in the valence band. [3 marks]
(iii). Holes positively charged charge carriers cannot exist in the conduction band of an element
like silicon. Justify this statement. [3 marks]
The electrons in the valence band are involved in chemical bonding and are typically tightly
bound to their respective atoms. They are not free to move around the solid because they require
a significant amount of energy to be excited to the conduction band. This energy can be provided
by thermal energy, light, or other means, but typically it is too high for the electrons to overcome
without external assistance. On the other hand, the electrons in the conduction band have energy
levels that are higher than those in the valence band and are not involved in bonding between
atoms. They are able to move freely throughout the solid because they are not tightly bound to
any specific atom. The presence of these free electrons is what makes a material conductive.
(iv). The density of copper is 8.93×10³kg/m³. Calculate the number of free electrons per cubic
meter and hence their drift velocity when a current is flowing whose density is 1 A/cm².
Exercise
Magnesium has one more electron than sodium, that is its 3S band is full. Explain why Mg is
also an electrical conductor. [2 marks]
Magnesium is an electrical conductor because it has free electrons in its conduction band.
Although magnesium has one more electron than sodium, the additional electron is in the 3p
orbital, which is higher in energy and not involved in metallic bonding. In the solid state,
magnesium atoms lose their outermost two electrons to form Mg2+ ions, which are surrounded
by a sea of delocalized electrons that are free to move throughout the material. These delocalized
electrons come from the 3s band, which is full in magnesium. When the outermost electrons are
removed, the 3s band becomes partially filled, with some of the electrons occupying higher
energy levels in the conduction band. These free electrons are able to move freely through the
material, conducting electricity
2. An intrinsic semiconductor has a conductivity of 390Ω‫־‬¹cm‫־‬¹ at 5⁰C and 1010 Ω‫־‬¹m‫־‬¹ at 25⁰C.
(i). What is the size of the energy gap in eV?
(ii). What is the conductivity at 15℃? [10 marks]
3. Explain what is a superconductor. State at least two practical applications of superconductors.
[4 marks]
A superconductor is a material that exhibits zero electrical resistance and expels magnetic fields
when cooled below a certain critical temperature.
1. Magnetic Resonance Imaging (MRI): Superconducting magnets are used in MRI machines to
produce a strong magnetic field that aligns the protons in a patient's body.
2. Electric power transmission: Since superconductors have zero electrical resistance, they can
carry much higher currents without losing energy to heat.
4. The electrical conductivity of Nickel decreases when a small amount of copper is added. This
is despite the fact that copper has a superior current conductivity than Nickel. Explain why the
resistivity increases in the Nickel-Copper alloy. [3 marks].
In a pure metal like nickel, the free electrons are able to move relatively easily through the
material, resulting in high electrical conductivity. However, when a small amount of copper is
added to the nickel, the arrangement of atoms in the alloy changes, leading to the scattering of
free electrons and increased resistance to their flow… Another factor that can contribute to the
increased resistivity of the alloy is the difference in electronic structure between nickel and
copper. Nickel has a partially filled d-band, while copper has a partially filled s-band. When
copper atoms are added to the nickel lattice, they can create energy levels in the conduction band
that are less favorable for electron transport, leading to a reduction in electrical conductivity.
5. Discuss electric current semi-conduction in amorphous materials. [3 marks].
In amorphous semiconductors, the lack of long-range order makes the band structure more
complex, and as a result, the electrical conductivity arises from the movement of charge carriers
in localized states known as defect states. The conductivity of amorphous semiconductors is
highly dependent on the density of defect states within the band gap, as well as the mobility of
the charge carriers.
6. (a). Explain why a semi-conductor exhibits a negative temperature coefficient of resistivity. [2
marks] A semiconductor exhibits a negative temperature coefficient of resistivity because the
number of charge carriers in the material increases with temperature
(b). Discuss why silver is the best metallic conductor of electricity. [2 marks] silver's atomic
structure, high electron mobility, low resistivity, high thermal conductivity, chemical stability,
ductility, and availability make it the best metallic conductor of electricity.
(c). Explain how conductivity modulation in semiconductors is used in thermistors and
photoconductors. [4 marks] Thermistors are made from semiconducting materials that exhibit a
strong temperature dependence of their conductivity. This property is achieved by doping the
semiconductor material with impurities that create defect states in the band gap. At low
temperatures, the impurities trap some of the charge carriers, reducing the conductivity of the
material. Photoconductors are made from semiconducting materials that are doped with
impurities that have an energy level within the band gap that is close to the conduction band.
When light is absorbed by the semiconductor material, electrons in the valence band are excited
to the conduction band, creating additional free charge carriers that increase the conductivity of
the material. The increase in conductivity can be used to detect the intensity of the incident light.
7. List four main causes of resistivity in metals. [4 marks]
1. Scattering of electrons: When an electric current flows through a metal, the electrons carrying
the current collide with impurities, defects, or lattice vibrations of the metal atoms, causing them
to scatter. These collisions cause a deviation in the path of electrons, which results in a decrease
in the average drift velocity of electrons and hence an increase in resistivity.
2. Electron-phonon interaction: Electrons in a metal can also interact with phonons, which are
lattice vibrations of the metal atoms. This interaction results in a loss of energy and momentum,
leading to scattering of electrons and an increase in resistivity.
3. Temperature: The resistivity of a metal increases with an increase in temperature. This is
because, at higher temperatures, there is an increase in lattice vibrations, which leads to an
increase in the number of scattering events and hence an increase in resistivity.
4. Grain boundaries: Metals are made up of grains, and at the boundaries between the grains,
there is a mismatch in the crystal structure of the metal atoms, which can lead to scattering of
electrons and hence an increase in resistivity.
8. The resistivity of pure silicon is 2.3×10³ Ω at 27℃. Find its conductivity at 200⁰C. Assume
that its energy gap is 1.1eV and k=8.65×10‫־‬⁵ eV/Kg answer1.04Ω‫־‬¹ m‫־‬¹ [5 marks]
Chapter Two
Exercise
Question One
(i). Define what is a dielectric material? A dielectric material is a type of insulating material that
does not conduct electricity easily.
(ii) Explain how we quantify the electrical behavior of materials. Conductivity is a measure of
how well a material conducts electricity and is defined as the inverse of resistivity. Resistivity is
a measure of the resistance of a material to the flow of electrical current and is the inverse of
conductivity. Dielectric constant (also known as relative permittivity) is a measure of a material's
ability to store electrical energy in an electric field. Magnetic permeability is a measure of a
material's ability to conduct magnetic fields.
Question Two
A 0.1 𝜇𝐹 capacitor is connected across 400 volts d.c. supply. If the relative dielectric constant of
the material is 4, calculate
(i). the energy stored in the capacitor and in polarizing the dielectric material (0.008 & 0.006 J).
1
𝛦 = 2 × 𝐶 × 𝑉2
1
𝐸 = 2 × 0.1 × 10−6 × 4002 = 0.008𝐽
1
1
E = 2 × 𝐶 × 𝑉 2 × (𝜀𝑟 − 1)
𝐸 = 2 × 0.1 × 10−6 × 4002 × (4 − 1) = 0.006𝐽
(ii). Repeat the calculations if the relative dielectric constant of the material is 91 (0.008 and
1
𝐸 = 2 × 0.1 × 10−6 × 4002 = 0.008𝐽
0.0072 J).
𝐸=
1
× 0.1 × 10−6 × 4002 × (91 − 1) = 0.0072𝐽
2
Question Three
A parallel plate capacitor has an area of 12cm² with a separation of 0.1mm. The space is filled
with Teflon. The real part of the complex relative dielectric constant is 2.1 and the loss tangent is
2×10−4, calculate the capacitance and the equivalent parallel loss resistance at a frequency of
(i). 1MHz (223pF, R =3.57MΩ, R= 35.7kΩ
𝐶=
𝜀𝑟 𝜀0 𝐴
𝑑
𝜀𝑟 = 𝜀𝑟 − 𝑗 tan 𝛿, 𝑎𝑡 1𝑀𝐻𝑧
𝐶=
(2.1−𝑗2×10−4 )𝜀0 (12×10−4 𝑚2 )
0.1×10−3
≈ 223𝑝𝐹
The equivalent parallel loss resistance at 1 MHz is given by:
𝑅=
1
1
=
≈ 35.7𝑘𝛺
6
(2𝜋𝑓𝐶 tan 𝛿) (2𝜋 × 100 × 10 𝐻𝑧 × 223 × 10−12 𝐹 × 2 × 10−4 )
1
(ii). 100MHz (tan δ = 1/(ω𝐶𝑅𝑎𝑐 ) tan 𝛿 = 𝜔𝐶𝑅 assume Ԑᵣ and tan δ are constants at the two
𝑎𝑐
frequencies.
(2.1 − 𝑗2 × 10−4 )𝜀0 (12 × 10−4 𝑚2 )
𝐶=
≈ 223𝑝𝐹
0.1 × 10−3
The equivalent parallel loss resistance at 100 MHz is given by:
𝑅=
1
1
=
≈ 35.7𝑘𝛺
6
(2𝜋𝑓𝐶 tan 𝛿) (2𝜋 × 100 × 10 𝐻𝑧 × 223 × 10−12 𝐹 × 2 × 10−4 )
Exercise
Question One
(a). A capacitor of 1μF has effective area of 480 cm² and it uses a material with relative dielectric
constant of 3. If a potential difference of 40V is applied to the capacitor, calculate the field
strength and the total dipole moment induced in the dielectric material. [4 marks]
(b). Discuss the following types of polarization in dielectric materials. [6 marks]
(i). Ionic polarization –occurs in materials that have ionic bonds, in these materials, the atoms are
held together by electrostatic forces between positively charged ions (cations) and negatively
charged ions (anions). When an electric field is applied, the ions experience relative
displacements, causing a net dipole moment to be induced in the material.
(ii). Electronic Polarization, occurs in materials that have covalent bonds, in these materials, the
electrons are shared between atoms in the bond. When an electric field is applied, the electrons
are displaced from their equilibrium positions, causing a net dipole moment to be induced in the
material.
(iii). Orientational Polarization, occurs in materials that have polar molecules, in these materials,
the molecules have a permanent dipole moment due to the unequal distribution of charges within
the molecule. When an electric field is applied, the polar molecules align themselves with the
field, causing a net dipole moment to be induced in the material
Question Two
a). Give the classification of ferroelectric materials and state the characteristic properties of each
type [8 marks]
] 1. Perovskite-type ferroelectrics: These materials have a perovskite crystal structure, which
consists of a central cation (usually a metal) surrounded by oxygen octahedra. They exhibit a
spontaneous polarization due to the displacement of the central cation from the center of the
octahedra. Examples of perovskite-type ferroelectrics include, barium titanate (BaTiO3), lead
zirconate titanate (PZT), and strontium bismuth tantalate (SBT).
Perovskite-type ferroelectrics: They have a high dielectric constant, large piezoelectric
coefficients, and a high Curie temperature (the temperature at which the spontaneous
polarization disappears). They are widely used in electronic devices, such as capacitors,
transducers, and actuators.
2. Layer-type ferroelectrics: These materials have a layered crystal structure, with the polar axes
perpendicular to the layers. They exhibit a spontaneous polarization due to the displacement of
ions within the layers. Examples of layer-type ferroelectrics include lead titanate (PbTiO3), lead
zirconate (PbZrO3), and lithium niobate (LiNbO3).
Layer-type ferroelectrics: They have a high electromechanical coupling coefficient, which makes
them suitable for use in ultrasound transducers and other medical applications. They also exhibit
nonlinear optical properties, which make them useful for frequency doubling and other optical
applications.
3. Polymeric ferroelectrics: These materials are organic polymers with polar functional groups,
such as polyvinylidene fluoride (PVDF) and its copolymers. They exhibit a spontaneous
polarization due to the alignment of the dipole moments of the polar groups.
Polymeric ferroelectrics: They are flexible, lightweight, and biocompatible, which makes them
suitable for use in sensors, energy harvesters, and biomedical devices. They also exhibit a large
pyroelectric coefficient, which makes them useful for thermal sensing and imaging applications.
(b). A parallel plate capacitor has an area of 8cm² with a separation of 0.08mm. The space is
filled with polystyrene. The real part of the complex relative dielectric constant is 2.56 and the
loss tangent is 0.7×10‫־‬⁴ at a frequency of 1MHz. Calculate the capacitance, the equivalent
parallel loss resistance and dielectric loss. Given E= 100𝑐𝑜𝑠2𝜋𝑓𝑡 V/m [8 marks]
Exercise
(a). Discuss the following materials and state their practical engineering applications: [9 marks]
Thermoelectric materials like thermocouple materials. Thermoelectric materials are used to
convert heat directly into electrical energy, they are made from two different metals that are
joined at one end. When there is a temperature difference between the two ends, a voltage is
generated across the junction due to the See beck effect.; Uses include,1. Power generation from
waste heat: 2. Temperature measurement 3. Cooling applications.
Magneto- electric materials. Magneto-electric materials are materials that exhibit both magnetic
and electric properties. They have the ability to convert magnetic fields into electrical signals.
Uses include,1. Magnetic sensors 2. Data storage 3. Energy harvesting.
Piezo-magnetic materials. Piezo-magnetic materials are materials that exhibit both piezoelectric
and magnetic properties. They have the ability to convert mechanical energy into electrical
energy.
(b). Differentiate piezo-electric effect from electrostriction effect. [4 marks]
Piezoelectric effect refers to the generation of an electric charge in response to mechanical stress
applied to a material, while electrostriction effect refers to the deformation of a material in
response to an applied electric field.
(c). With clear examples discuss the various applications of piezo electrics. [8 marks]
1.Sensors: Piezoelectric materials are commonly used in sensors such as pressure sensors,
accelerometers, and vibration sensors.
2. Actuators: Piezoelectric materials can be used as actuators in a variety of applications,
including micro-robotics, precision machining, and medical devices.
3. Energy harvesting: Piezoelectric materials can be used to harvest energy from ambient
vibrations and mechanical movements which is used to power small electronic devices such as
sensors and wireless communication systems.
4. Medical applications: Piezoelectric materials are commonly used in medical devices such as
ultrasound machines, which use piezoelectric transducers to generate and receive sound waves
for imaging.
Chapter Three
Exercise
What is magnetic anisotropy energy?
Magnetic anisotropy energy (MAE) refers to the energy required to change the direction of
magnetization in a magnetic material from its preferred orientation or axis. In other words, it is a
measure of the degree to which a magnetic material prefers to have its magnetization oriented
along a certain direction, rather than in any other direction.
Explain why magnetostriction is anisotropic.
Magnetostriction is anisotropic because the crystal structure of the magnetic material is also
anisotropic. In other words, the crystal structure of the material is not uniform in all directions.
This anisotropy affects the response of the material to an applied magnetic field.
Elaborate the factors that affect the magnitude of coercive force and hence the remanence flux
density.
Material properties: The coercive force and remanence flux density are dependent on the type of
material being used. Materials with high magnetic anisotropy, such as rare earth metals, have
high coercive force and remanence flux density, while materials with low magnetic anisotropy,
such as iron, have low coercive force and remanence flux density.
Shape and size of the magnet: The shape and size of a magnet can affect the magnitude of
coercive force and remanence flux density. Thin and elongated magnets have lower coercive
force and remanence flux density than thick and short magnets.
Temperature: Temperature has a significant effect on the coercive force and remanence flux
density of ferromagnetic materials. As temperature increases, the coercive force decreases,
leading to a decrease in remanence flux density. This is due to the increased thermal energy that
destabilizes the magnetic domains in the material.
Magnetization history: The previous magnetization history of a ferromagnetic material can affect
its coercive force and remanence flux density. If a material has been previously magnetized in
the opposite direction, it may require a higher coercive force to demagnetize it and produce a
lower remanence flux density.
Applied magnetic field strength: The magnitude of the magnetic field strength applied during the
magnetization process affects the coercive force and remanence flux density of the material.
Higher magnetic field strengths during magnetization result in higher coercive force and
remanence flux density.
Elucidate clearly the reasons why iron crystal expands when magnetized in the easy direction of
magnetization and contracts when magnetized in the hard direction of magnetization.
When iron is magnetized in the easy direction of magnetization, the magnetic domains within the
crystal align parallel to the direction of the magnetic field. This alignment of the domains causes
the lattice structure to expand due to the repulsion between the aligned domains, leading to an
increase in the overall size of the crystal.
On the other hand, when iron is magnetized in the hard direction of magnetization, the magnetic
domains align perpendicular to the direction of the magnetic field. This alignment causes a
contraction of the lattice structure due to the attraction between the aligned domains. This
contraction leads to a decrease in the overall size of the crystal.
Discuss the procedure that would you recommend for making the materials required for
permanent magnets.
Material selection: The first step is to select the raw materials that will be used to create the
permanent magnet. The most common materials used for permanent magnets are iron, cobalt,
and nickel, as well as their alloys.
Melting: The selected raw materials are melted together in a furnace at high temperatures. The
precise temperature and duration of melting can vary depending on the specific materials being
used.
Alloying: Once the materials have melted, other elements are added to the mixture to create the
desired alloy. For example, adding aluminum to an iron-cobalt-nickel alloy can increase its
magnetic strength.
Casting: The molten alloy is then poured into a mold to create the desired shape of the permanent
magnet. This can be done using a variety of techniques, such as die casting, injection molding, or
powder metallurgy.
Cooling: The cast magnet is cooled slowly to room temperature to ensure that it retains its
magnetic properties.
Grinding and shaping: The cooled magnet is ground and shaped to the desired final dimensions.
Magnetization: The final step in the process is to magnetize the permanent magnet. This can be
done by exposing the magnet to a strong magnetic field.
Overall, making materials for permanent magnets is a complex process that requires careful
selection of raw materials, precise melting and alloying techniques, and careful cooling and
magnetization steps to ensure that the final product has the desired magnetic properties.
What causes humming noise in a power transformer? How can the hysteresis and eddy current
losses in such devices be minimized?
The humming noise in a power transformer is caused by the magnetic core’s vibration, which is
due to the alternating magnetic field generated by the transformer’s current. The magnetic core is
constructed of thin laminated sheets of steel or iron, which can vibrate at the AC frequency and
produce an audible hum.
Hysteresis loss occurs when the transformer core’s magnetic field is continuously changing
direction and magnitude, causing the magnetic domains to align and then realign with the
changing magnetic field, which results in energy loss. Eddy current loss is caused by the
circulating currents generated within the transformer core by the changing magnetic field.
To minimize hysteresis and eddy current losses in power transformers, several techniques can be
employed. One approach is to use laminated cores made of high-quality steel or iron, which are
specifically designed to reduce the losses associated with hysteresis and eddy currents. Another
technique is to increase the core’s thickness, as thicker cores can reduce eddy current losses.
Furthermore, the use of soft magnetic materials like ferrites, amorphous alloys or nanocrystalline
cores can significantly reduce the hysteresis and eddy current losses in a transformer, thus
minimizing the humming noise. Additionally, proper design and construction techniques such as
using interleaved windings, increasing the number of turns, and reducing the air gap between the
windings and the core can also minimize transformer losses and humming noise.
What is Curie temperature? Give its value for the ferromagnetic elements and hence explain why
these elements become paramagnetic at and above this temperature.
Curie temperature, also known as the Curie point, is the temperature at which a ferromagnetic
material loses its ferromagnetic properties and becomes paramagnetic. Above this temperature,
the thermal energy overcomes the magnetic interactions, causing the material to lose its
magnetization.
The value of Curie temperature varies depending on the specific ferromagnetic material. For
example, the Curie temperature of iron is 770℃, while that of cobalt is 1121℃, and nickel is
358℃.
At temperatures below the Curie temperature, ferromagnetic materials have spontaneously
aligned magnetic domains, resulting in a strong net magnetic moment. However, when the
temperature is raised above the Curie temperature, the thermal energy becomes sufficient to
disrupt the alignment of the magnetic domains, and the net magnetic moment decreases rapidly.
As a result, ferromagnetic materials lose their ferromagnetic properties above the Curie
temperature and become paramagnetic, which means that they are weakly attracted to an external
magnetic field but do not retain any magnetization once the field is removed.
In summary, the Curie temperature is the temperature at which ferromagnetic materials lose their
ferromagnetic properties and become paramagnetic due to the thermal energy overcoming the
magnetic interactions between magnetic moments.
8. The rectangular hysteresis curve of a material has a coercity of 50Am ‫־‬¹ and a remanence of
0.5T. Estimate the hysteresis loss per cubic meter of the material that is dissipated when the
material is driven around one complete hysteresis cycle. If this material is used in a toroidal
inductor core of mean circumference 0.05m with a cross-sectional area of 0.25×10‫־‬⁴m², calculate
the hysteresis power loss at a frequency of 60Hz. Is this the total power loss in the inductor at
this frequency? What materials properties would you take into account in selecting a material as
an inductor core?
Exercise
What is magnetostriction? State its cause and list one advantage and one disadvantage of
magnetostriction. [5 marks]
Magnetostriction is a phenomenon in which a magnetic material experiences a change in its
dimensions when subjected to a magnetic field. This change in dimensions is typically in the
form of a deformation or a strain in the material.
The cause of magnetostriction is the interaction between the magnetic field and the magnetic
domains within the material. As the magnetic field changes, the magnetic domains align
themselves with the field, resulting in a change in the dimensions of the material.
One advantage of magnetostriction is that it can be used to convert magnetic energy into
mechanical energy. This property has led to its use in a variety of applications such as sensors,
actuators, and transducers.
One disadvantage of magnetostriction is that it can lead to mechanical stress and fatigue in the
material. This can result in degradation of the material over time, especially if it is subjected to
repeated cycles of magnetic fields. Additionally, the deformation caused by magnetostriction can
lead to unwanted noise and vibrations in certain applications.
Explain why power transformers cores are made from cold rolled grain-oriented silicon steel
alloy. [7 marks]
Power transformers are used to transfer electrical energy from one circuit to another by means of
electromagnetic induction. The transformer core is an Important component of the transformer
that helps to transfer the energy efficiently from the primary to the secondary winding. The core
is typically made of laminated sheets of steel, which are stacked and joined together to form a
closed magnetic circuit.
Cold rolled grain-oriented silicon steel alloy is the preferred material for transformer cores
because it exhibits very low hysteresis loss and eddy current loss, which are the two main
sources of energy loss in a transformer. These losses occur due to the magnetic properties of the
core material, and they cause the core to heat up, resulting in energy loss and reduced efficiency.
The use of grain-oriented silicon steel alloy in transformer cores helps to minimize these losses
by reducing the hysteresis and eddy current losses. The grain-oriented silicon steel alloy is
produced using a special manufacturing process that aligns the grains of the steel in a specific
direction, resulting in a very low magnetic reluctance and high magnetic permeability. This
ensures that the magnetic field is efficiently transferred through the core, resulting in minimal
energy loss and high efficiency.
Note, the use of cold rolled grain-oriented silicon steel alloy in transformer cores results in a
more efficient and reliable transformer, with reduced energy loss and improved performance.
3. The following are typical metallic alloys and ceramic materials used in magnetic applications;
discuss their applications and how their properties and behavior can be enhanced. [21 marks]
Magnetic Metals: Magnetic metals such as iron, nickel, and cobalt are used extensively in
magnetic applications. They exhibit strong ferromagnetism, which makes them ideal for use in
permanent magnets, electromagnets, and magnetic storage devices. Their magnetic properties
can be enhanced by alloying them with other metals such as aluminum, copper, and titanium.
These alloys exhibit improved magnetic properties and higher resistance to corrosion. They are
widely used in the automotive, electronics, and aerospace industries.
Iron-Nickel Alloys: Iron-nickel alloys, also known as invar, exhibit low thermal expansion and
high magnetic permeability. They are used in precision instruments, such as clocks and watches,
and in the aerospace industry for manufacturing parts that require stability at different
temperatures. The addition of chromium and molybdenum to iron-nickel alloys improves their
resistance to corrosion, making them suitable for use in marine applications.
Silicon Iron: Silicon iron, also known as electrical steel, is a magnetic material used in the
production of transformers, generators, and electric motors. The addition of silicon to iron
reduces the hysteresis loss and eddy current losses, resulting in improved magnetic properties.
The grain-oriented electrical steel exhibits anisotropic magnetic behavior, which allows it to be
used in the production of electrical transformers.
Composite Magnets: Composite magnets are made by combining magnetic particles with a nonmagnetic matrix, such as plastic or epoxy resin. They exhibit improved magnetic properties, such
as high coercivity, and are used in various magnetic applications, including sensors, actuators,
and magnetic bearings. The magnetic properties of composite magnets can be enhanced by
optimizing the particle size, shape, and distribution.
Metallic Glasses: Metallic glasses are amorphous materials made by rapidly cooling molten
metal. They have unique magnetic properties that make them useful in a variety of applications,
such as magnetic sensors, transformers, and motors. Their properties can be enhanced by
controlling the cooling rate during the manufacturing process. The faster the cooling rate, the
more disordered the atoms become, leading to increased magnetic properties. Metallic glasses
can also be doped with other elements to enhance their magnetic properties, such as adding iron
to a nickel-based metallic glass to create a soft magnetic material.
Magnetic Tape: Magnetic tape is a thin strip of plastic coated with a magnetic material, typically
iron oxide. It is used for data storage in devices such as tape drives, cassette tapes, and VHS
tapes. The properties of magnetic tape can be enhanced by increasing the coercivity, which is the
measure of a material’s resistance to demagnetization. This is achieved by increasing the
magnetic particle size and decreasing the spacing between the particles on the tape. Additionally,
magnetic tape can be made more durable by adding a protective coating, such as a polymer or
metal layer.
Complex Metallic Alloys for Permanent Magnets: Complex metallic alloys, such as neodymiumiron-boron (NdFeB) and samarium-cobalt (SmCo), are used in the production of permanent
magnets. These magnets are used in a variety of applications, such as electric motors, generators,
and MRI machines. The properties of complex metallic alloys can be enhanced by altering the
alloy composition and microstructure. For example, adding small amounts of dysprosium to a
NdFeB magnet can improve its high-temperature performance. Additionally, the microstructure
of the magnet can be controlled during the manufacturing process to optimize the magnetic
properties, such as grain size, crystal structure, and magnetic domain structure.
Exercise
(i). Elaborate why Diamagnetic materials are repelled by permanent magnets. [2 marks]
When a diamagnetic material is placed near a permanent magnet, the magnetic field from the
magnet induces a small magnetic field in the opposite direction within the diamagnetic material.
This induced magnetic field opposes the magnetic field from the permanent magnet and results
in a repulsive force between the two objects.
(ii). Discuss how a high coercivity is achieved in permanent magnets. [3 marks]
A high coercivity is achieved in permanent magnets by using materials with a high magnetic
anisotropy, such as neodymium-iron-boron (NdFeB) or samarium-cobalt (SmCo) alloys. These
materials have a crystalline structure that aligns the magnetic moments of the individual atoms in
a preferred direction, resulting in a strong overall magnetic field.
Additionally, the magnets are typically processed in a way that aligns their crystal structure and
magnetic domains, which enhances their coercivity. For example, NdFeB magnets are usually
manufactured using a powder metallurgy process, where the powder is compacted under a strong
magnetic field, and then sintered at high temperatures to form a dense, solid magnet.
Finally, post-processing techniques such as heat treatment and magnetization can further increase
the coercivity of the permanent magnet. These steps are designed to optimize the alignment of
the magnetic domains within the magnet, resulting in a stronger and more stable magnetic field.
(iii). What is magnetic susceptibility? List two factors which may influence the magnetic
susceptibility of a magnetic material. [3 marks]
Magnetic susceptibility is a measure of the degree of magnetization that a material will
experience when subjected to a magnetic field. It is defined as the ratio of the magnetization of
the material to the applied magnetic field.
Two factors that can influence the magnetic susceptibility of a material are:
Atomic structure: The magnetic properties of a material are heavily influenced by its atomic
structure, specifically the arrangement of electrons in the outermost energy level. Materials with
unpaired electrons in their outermost energy level tend to exhibit higher magnetic susceptibility.
Temperature: The temperature of a material can also affects its magnetic susceptibility. At higher
temperatures, the thermal energy can cause the magnetic domains in the material to become
disordered, leading to a decrease in the material’s magnetic susceptibility.
(iv). A 200 turns 40m coil is carrying a current of 20A. It is wound on a material with a relative
permeability of 5000. Calculate the magnetization (M) and inductance (B) in the material. [4
marks]
To calculate the magnetization and inductance of the given coil, we can use the following
equations:
𝑁×𝐼
Magnetization (M) =
Inductance(L) =
𝑙
(𝜇×𝑁 2 ×A)
𝑙
Where N is the number of turns, I is the current, l is the length of the coil, A is the cross-sectional
area of the coil, and µ is the permeability of the material.
Given data:
Number of turns (N) = 200
Current (I) = 20 A
Length of coil (l) = 40 m
Relative permeability (µ) = 5000
Cross-sectional area (A) of the coil is not given, so we can assume a circular cross-section with a
radius of 0.1 m (since the length is 40 m and the coil is wound in a single layer, the coil diameter
is 0.2 m).
Cross-sectional area (𝐴) = 𝜋𝑟 2 = 𝜋 × 0.12 = 0.0314 m²
Now we can calculate the magnetization:
Magnetization (M) =
𝑁×𝐼
𝑙
=
200×20
40
= 100 A/m
Next, we can calculate the inductance:
Inductance (L) =
(𝜇×𝑁 2 ×A)
𝑙
=
5000×2002 ×0.0314
40
= 7.85 H
Therefore, the magnetization (M) in the material is 100 A/m and the inductance (L) of the coil is
7.85 H.
(v). Define with an example the following materials and discuss their main applications in
engineering. [12 marks]
Dielectric Materials: Dielectric materials are those materials that have a very low electrical
conductivity and are commonly used in electrical insulation applications. They are used to store
electrical energy in capacitors and to isolate conductive materials from each other. A common
example of a dielectric material is polyethylene, which is used in the production of cables and
electrical wires.
Electrical Conducting Materials: Electrical conducting materials are those materials that can
carry an electrical current with a very low resistance. They are used to transfer electrical energy
from one point to another and to make electrical connections. Examples of electrical conducting
materials include copper, aluminum, and gold, which are widely used in electrical wiring, circuit
boards, and electronic components.
Magnetic Materials: Magnetic materials are those materials that possess magnetic properties and
are used in various applications related to magnetism. They are used in the production of
electrical motors, generators, and transformers. Examples of magnetic materials include iron,
cobalt, and nickel, which are widely used in the production of permanent magnets, and soft
magnetic materials such as ferrite, which are used in the production of inductors and
transformers.
Electrical Insulating Materials: Electrical insulating materials are those materials that can prevent
the flow of electrical current through them. They are used to provide electrical insulation
between conductive materials and to protect electrical components from damage. Examples of
electrical insulating materials include rubber, PVC, and silicone, which are widely used in the
production of electrical wires, cables, and electrical equipment.
Chapter Four
Exercise
Discuss the remaining types of encapsulation methods.
Polymer encapsulation: Polymer encapsulation involves using a polymer material to encapsulate
another material. Polymers can be used to encapsulate a variety of materials such as drugs,
chemicals, and biological substances. Polymer encapsulation is a popular method because it
provides a protective barrier and controls the release of the encapsulated material.
Microencapsulation: Microencapsulation involves enclosing small particles or droplets of a
substance within a protective shell. Microencapsulation is commonly used to encapsulate food
ingredients, pharmaceuticals, and fragrances. The shell material can be made from various
materials such as polymers, lipids, or proteins.
Nanoparticle encapsulation: Nanoparticle encapsulation involves using nanoparticles to
encapsulate other materials. Nanoparticles have unique properties such as high surface area and
high reactivity, making them useful for encapsulating and delivering drugs and other therapeutic
agents.
Liposome encapsulation: Liposome encapsulation involves using a lipid bilayer to encapsulate a
substance. Liposomes are commonly used to deliver drugs and other therapeutic agents because
they can be designed to target specific tissues or organs in the body.
Inorganic encapsulation: Inorganic encapsulation involves using inorganic materials such as
silica or metal oxides to encapsulate other materials. Inorganic encapsulation is often used to
protect sensitive materials from environmental factors such as temperature, humidity, and light.
Overall, encapsulation methods are widely used in materials science to protect and control the
release of sensitive materials.